Problem: The fifth term of a geometric sequence of positive numbers is $11$ and the eleventh term is $5$. What is the eighth term of the sequence? Express your answer in simplest radical form.  [asy]
size(150); defaultpen(linewidth(2));
real loc = 0;
for(int i = 0; i < 11; ++i) {

if(i == 4)

label("$\mathbf{\mathit{11}}$",(loc,0),(0.8,1.2),fontsize(14));

if(i == 10)

label("$\mathbf{\mathit{5}}$",(loc,0),(1.2,1.2),fontsize(14));

fill(box((loc,0),(loc+1,0.15)));

loc += 4/3;
}
[/asy]
Explanation: Let $r$ be the common ratio of the geometric sequence. Then, the eighth term of the sequence is equal to $11r^3$, and the eleventh term of the sequence is equal to $11r^6 = 5$. From the second equation, it follows that $r^6 = \frac{5}{11} \Longrightarrow r^3 = \sqrt{\frac{5}{11}}$. Thus, $11r^3 = 11 \cdot \sqrt{\frac{5}{11}} = \sqrt{\frac{11^2 \cdot 5}{11}} = \boxed{\sqrt{55}}$.

Alternatively, since the eighth term is the middle term between the fifth term and the eleventh term, it follows that the eighth term is the geometric mean of the fifth and eleventh terms.